Therefore c x y Ax b y Lizhi Wang lzwangiastateedu IE 534 Linear Programming

Therefore c x y ax b y lizhi wang lzwangiastateedu ie

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Therefore, c > x y > Ax b > y . Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 11 / 21
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Outline 1 Review 2 Complementary slackness 3 Weak duality 4 Strong duality 5 Farkas lemma 6 Primal-dual possibilities Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 12 / 21
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Strong duality Theorem (Strong duality) If x and y are optimal solutions to max { c > x : Ax b, x 0 } and min { b > y : A > y c, y 0 } , respectively, then c > x = b > y. Proof. Since x and y are optimal, by complementary slackness, we have 0 = ( b - Ax ) > y = b > y - ( Ax ) > y = b > y - y > Ax, and 0 = ( A > y - c ) > x = y > Ax - c > x. Therefore, c > x = y > Ax = b > y . Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 13 / 21
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Outline 1 Review 2 Complementary slackness 3 Weak duality 4 Strong duality 5 Farkas lemma 6 Primal-dual possibilities Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 14 / 21
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Rethink unboundedness Consider the standard from LP max x { c > x : Ax b, x 0 } . An LP being unbounded means two things 1 a feasible solution x 0 , which satisfies Ax 0 b, x 0 0 2 a direction Δ x , which leads lim λ →∞ c > ( x 0 + λ Δ x ) → ∞ without making ( x 0 + λ Δ x ) infeasible This direction Δ x is called an extreme ray . It satisfies: A Δ x 0 , Δ x 0 , and c > Δ x > 0 . Consider the dual LP min y { b > y : A > y c, y 0 } . The dual extreme ray Δ y satisfies: A > Δ y 0 , Δ y 0 , and b > Δ y < 0 . Lizhi Wang ([email protected]) IE 534 Linear Programming September 12, 2012 15 / 21
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Farkas lemma Theorem (Farkas lemma) Let A R m × n and b R m × 1 be a matrix and a vector, respectively. Then exactly one of the following two alternatives holds: (a) There exists some x 0 such that Ax b . (b) There exists some y 0 such that A > y 0 and b > y < 0 . Proof. (a) true (b) false: If (a) is true, then for any y 0 such that A > y 0 , we have b > y ( Ax ) > y = x > A > y 0 , which means that (b) is false.
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